A single force acts on a 5.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t - 4.0t 2 + 1.0t 3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 2.0 s.
find the position at at t=2 and t=0. The difference in those is the displacement.
Work= force times displacement.
So when t=0, x=0.
When t=2, x=-2
displacement= -2-0 = -2m
W=Fd
F=ma=(5)(9.8)=49
W=(49 N)(-2 m)
W= -98 J
Is this correct??
no>>>
find the third divetive then substute the a in F=ma< and you will get the force then find the destanse moved then
W=F x
To find the position at t=2 and t=0, you can simply substitute the values of t into the given equation for x.
When t=0,
x = 3.0(0) - 4.0(0^2) + 1.0(0^3)
= 0 - 0 + 0
= 0
So, at t=0, the position (x) of the object is 0 meters.
When t=2,
x = 3.0(2) - 4.0(2^2) + 1.0(2^3)
= 6.0 - 4.0(4) + 1.0(8)
= 6.0 - 16.0 + 8.0
= -2.0
So, at t=2, the position (x) of the object is -2.0 meters.
The displacement is the difference between the positions at t=2 and t=0:
Displacement = x(t=2) - x(t=0)
= (-2.0 meters) - (0 meters)
= -2.0 meters
Now, you can calculate the work done on the object by the force using the formula:
Work = Force x Displacement
Given that the mass of the object is 5.0 kg, and the force required can be calculated as F = ma = (5.0 kg)(9.8 m/s^2) = 49 N:
Work = (49 N)(-2.0 m)
= -98 J
So, the work done on the object by the force from t=0 to t=2.0 s is -98 Joules (J).